Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{-7y - 28}{y + 6} \div \dfrac{y^2 + 10y}{y^2 + 16y + 60} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-7y - 28}{y + 6} \times \dfrac{y^2 + 16y + 60}{y^2 + 10y} $ First factor the quadratic. $t = \dfrac{-7y - 28}{y + 6} \times \dfrac{(y + 6)(y + 10)}{y^2 + 10y} $ Then factor out any other terms. $t = \dfrac{-7(y + 4)}{y + 6} \times \dfrac{(y + 6)(y + 10)}{y(y + 10)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -7(y + 4) \times (y + 6)(y + 10) } { (y + 6) \times y(y + 10) } $ $t = \dfrac{ -7(y + 4)(y + 6)(y + 10)}{ y(y + 6)(y + 10)} $ Notice that $(y + 10)$ and $(y + 6)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ -7(y + 4)\cancel{(y + 6)}(y + 10)}{ y\cancel{(y + 6)}(y + 10)} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $t = \dfrac{ -7(y + 4)\cancel{(y + 6)}\cancel{(y + 10)}}{ y\cancel{(y + 6)}\cancel{(y + 10)}} $ We are dividing by $y + 10$ , so $y + 10 \neq 0$ Therefore, $y \neq -10$ $t = \dfrac{-7(y + 4)}{y} ; \space y \neq -6 ; \space y \neq -10 $